The Tsiolkovsky rocket equation, or ideal rocket equation, describes the motion of vehicles that follow the basic principle of a rocket: a device that can apply acceleration to itself (a thrust) by expelling part of its mass with high speed and thereby move due to the conservation of momentum. The equation relates the delta-v (the maximum change of velocity of the rocket if no other external forces act) with the effective exhaust velocity and the initial and final mass of a rocket (or other reaction engine).

For any such maneuver (or journey involving a sequence of such maneuvers):

(The equation can also be written using the specific impulse instead of the effective exhaust velocity by applying the formula ve=Isp⋅g0{\displaystyle v_{\text{e}}=I_{\text{sp}}\cdot g_{0}} where Isp{\displaystyle I_{\text{sp}}} is the specific impulse expressed as a time period and g0{\displaystyle g_{0}} is standard gravity ≈ 9.8 m/s2.)

The equation is named after Russian scientist Konstantin Tsiolkovsky who independently derived it and published it in his 1903 work.[1] The equation had been derived earlier by the British mathematician William Moore in 1813.[2] The minister William Leitch who was a capable scientist also independently derived the fundamentals of rocketry in 1861.

This equation was independently derived by Konstantin Tsiolkovsky towards the end of the 19th century and is sometimes known under his name, but more often simply referred to as 'the rocket equation' (or sometimes the 'ideal rocket equation').

While the derivation of the rocket equation is a straightforward calculus exercise, Tsiolkovsky is honored as being the first to apply it to the question of whether rockets could achieve speeds necessary for space travel.

V+ΔV{\displaystyle V+\Delta V\,} is the velocity of the rocket at time t=Δt{\displaystyle t=\Delta t\,}

Ve{\displaystyle V_{e}\,} is the velocity of the mass added to the exhaust (and lost by the rocket) during time Δt{\displaystyle \Delta t\,}

m+Δm{\displaystyle m+\Delta m\,} is the mass of the rocket at time t=0

m{\displaystyle m\,} is the mass of the rocket at time t=Δt{\displaystyle t=\Delta t\,}

The velocity of the exhaust Ve{\displaystyle V_{e}} in the observer frame is related to the velocity of the exhaust in the rocket frame ve{\displaystyle v_{e}} by (since exhaust velocity is in the negative direction)

Ve=V−ve{\displaystyle V_{e}=V-v_{e}}

Solving yields:

P2−P1=mΔV−veΔm{\displaystyle P_{2}-P_{1}=m\Delta V-v_{e}\Delta m\,}

and, using dm=−Δm{\displaystyle dm=-\Delta m}, since ejecting a positive Δm{\displaystyle \Delta m} results in a decrease in mass,

where m0{\displaystyle m_{0}} is the initial total mass including propellant, m1{\displaystyle m_{1}} the final total mass, and ve{\displaystyle v_{e}} the velocity of the rocket exhaust with respect to the rocket (the specific impulse, or, if measured in time, that multiplied by gravity-on-Earth acceleration).

The value m0−m1{\displaystyle m_{0}-m_{1}} is the total mass of propellant expended, and hence:

ΔV{\displaystyle \Delta V\ } (delta v) is the integration over time of the magnitude of the acceleration produced by using the rocket engine (what would be the actual acceleration if external forces were absent). In free space, for the case of acceleration in the direction of the velocity, this is the increase of the speed. In the case of an acceleration in opposite direction (deceleration) it is the decrease of the speed. Of course gravity and drag also accelerate the vehicle, and they can add or subtract to the change in velocity experienced by the vehicle. Hence delta-v is not usually the actual change in speed or velocity of the vehicle.

If special relativity is taken into account, the following equation can be derived for a relativistic rocket,[3] with Δv{\displaystyle \Delta v} again standing for the rocket's final velocity (after burning off all its fuel and being reduced to a rest mass of m1{\displaystyle m_{1}}) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being m0{\displaystyle m_{0}} initially), and c{\displaystyle c} standing for the speed of light in a vacuum:

Imagine a rocket at rest in space with no forces exerted on it (Newton's First Law of Motion). However, as soon as its engine is started (clock set to 0) the rocket is expelling gas mass at a constant mass flow rate M (kg/s) and at exhaust velocity relative to the rocket ve (m/s). This creates a constant force propelling the rocket that is equal to M × ve. The mass of fuel the rocket initially has on board is equal to m0 - mf. It will therefore take a time that is equal to (m0 - mf)/M to burn all this fuel. Now, the rocket is subject to a constant force (M × ve), but at the same time its total weight is decreasing steadily because it's expelling gas. According to Newton's Second Law of Motion, this can have only one consequence; its acceleration is increasing steadily. To obtain the acceleration, the propelling force has to be divided by the rocket's total mass. So, the level of acceleration at any moment (t) after ignition and until the fuel runs out is given by;

Mvem0−(Mt).{\displaystyle ~{\frac {Mv_{e}}{m_{0}-(Mt)}}.}

Since the time it takes to burn the fuel is (m0 - mf)/M the acceleration reaches its maximum of

Mvemf{\displaystyle ~{\frac {Mv_{e}}{m_{f}}}}

the moment the last fuel is expelled. Since the exhaust velocity is related to the specific impulse in unit time as

Since speed is the definite integration of acceleration, and the integration has to start at ignition and end the moment the last propellant leaves the rocket, the following definite integral yields the speed at the moment the fuel runs out;

Delta-v is produced by reaction engines, such as rocket engines and is proportional to the thrust per unit mass, and burn time, and is used to determine the mass of propellant required for the given manoeuvre through the rocket equation.

For multiple manoeuvres, delta-v sums linearly.

For interplanetary missions delta-v is often plotted on a porkchop plot which displays the required mission delta-v as a function of launch date.

In aerospace engineering, the propellant mass fraction is the portion of a vehicle's mass which does not reach the destination, usually used as a measure of the vehicle's performance. In other words, the propellant mass fraction is the ratio between the propellant mass and the initial mass of the vehicle. In a spacecraft, the destination is usually an orbit, while for aircraft it is their landing location. A higher mass fraction represents less weight in a design. Another related measure is the payload fraction, which is the fraction of initial weight that is payload.

The rocket equation captures the essentials of rocket flight physics in a single short equation. It also holds true for rocket-like reaction vehicles whenever the effective exhaust velocity is constant, and can be summed or integrated when the effective exhaust velocity varies. The rocket equation only accounts for the reaction force from the rocket engine; it does not include other forces that may act on a rocket, such as aerodynamic or gravitational forces. As such, when using it to calculate the propellant requirement for launch from (or powered descent to) a planet with an atmosphere, the effects of these forces must be included in the delta-V requirement (see Examples below). In what has been called "the tyranny of the rocket equation", there is a limit to the amount of payload that the airship can carry, as higher amounts of propellant increment the overall weight, and thus also increase the fuel consumption.[5] The equation does not apply to non-rocket systems such as aerobraking, gun launches, space elevators, launch loops, or tether propulsion.

The rocket equation can be applied to orbital maneuvers in order to determine how much propellant is needed to change to a particular new orbit, or to find the new orbit as the result of a particular propellant burn. When applying to orbital maneuvers, one assumes an impulsive maneuver, in which the propellant is discharged and delta-v applied instantaneously. This assumption is relatively accurate for short-duration burns such as for mid-course corrections and orbital insertion maneuvers. As the burn duration increases, the result is less accurate due to the effect of gravity on the vehicle over the duration of the maneuver. For low-thrust, long duration propulsion, such as electric propulsion, more complicated analysis based on the propagation of the spacecraft's state vector and the integration of thrust are used to predict orbital motion.

Assume an exhaust velocity of 4,500 meters per second (15,000 ft/s) and a Δv{\displaystyle \Delta v} of 9,700 meters per second (32,000 ft/s) (Earth to LEO, including Δv{\displaystyle \Delta v} to overcome gravity and aerodynamic drag).

Single-stage-to-orbit rocket: 1−e−9.7/4.5{\displaystyle 1-e^{-9.7/4.5}} = 0.884, therefore 88.4% of the initial total mass has to be propellant. The remaining 11.6% is for the engines, the tank, and the payload. In the case of a space shuttle, it would also include the orbiter.

Two-stage-to-orbit: suppose that the first stage should provide a Δv{\displaystyle \Delta v} of 5,000 meters per second (16,000 ft/s); 1−e−5.0/4.5{\displaystyle 1-e^{-5.0/4.5}} = 0.671, therefore 67.1% of the initial total mass has to be propellant to the first stage. The remaining mass is 32.9%. After disposing of the first stage, a mass remains equal to this 32.9%, minus the mass of the tank and engines of the first stage. Assume that this is 8% of the initial total mass, then 24.9% remains. The second stage should provide a Δv{\displaystyle \Delta v} of 4,700 meters per second (15,000 ft/s); 1−e−4.7/4.5{\displaystyle 1-e^{-4.7/4.5}} = 0.648, therefore 64.8% of the remaining mass has to be propellant, which is 16.2%, and 8.7% remains for the tank and engines of the second stage, the payload, and in the case of a space shuttle, also the orbiter. Thus together 16.7% is available for all engines, the tanks, the payload, and the possible orbiter.

In the case of sequentially thrusting rocket stages, the equation applies for each stage, where for each stage the initial mass in the equation is the total mass of the rocket after discarding the previous stage, and the final mass in the equation is the total mass of the rocket just before discarding the stage concerned. For each stage the specific impulse may be different.

For example, if 80% of the mass of a rocket is the fuel of the first stage, and 10% is the dry mass of the first stage, and 10% is the remaining rocket, then

If the motor of a new stage is ignited before the previous stage has been discarded and the simultaneously working motors have a different specific impulse (as is often the case with solid rocket boosters and a liquid-fuel stage), the situation is more complicated.

When viewed as a variable-mass system, a rocket cannot be directly analyzed with Newton's second law of motion because the law is valid for constant-mass systems only.[6][7][8] It can cause confusion that the Tsiolkovsky rocket equation looks similar to the relativistic force equation F=dp/dt=mdv/dt+vdm/dt{\displaystyle F=dp/dt=m\;dv/dt+v\;dm/dt}. Using this formula with m(t){\displaystyle m(t)} as the varying mass of the rocket seems to derive Tsiolkovsky rocket equation, but this derivation is not correct. Notice that the effective exhaust velocityve{\displaystyle v_{e}} doesn't even appear in this formula.

^Plastino, Angel R.; Muzzio, Juan C. (1992). "On the use and abuse of Newton's second law for variable mass problems". Celestial Mechanics and Dynamical Astronomy. Netherlands: Kluwer Academic Publishers. 53 (3): 227–232. Bibcode:1992CeMDA..53..227P. doi:10.1007/BF00052611. ISSN0923-2958. "We may conclude emphasizing that Newton's second law is valid for constant mass only. When the mass varies due to accretion or ablation, [an alternate equation explicitly accounting for the changing mass] should be used."

^Halliday; Resnick. Physics. 1. p. 199. ISBN0-471-03710-9. It is important to note that we cannot derive a general expression for Newton's second law for variable mass systems by treating the mass in F = dP/dt = d(Mv) as a variable. [...] We can use F = dP/dt to analyze variable mass systems only if we apply it to an entire system of constant mass having parts among which there is an interchange of mass. [Emphasis as in the original]

^Kleppner, Daniel; Robert Kolenkow (1973). An Introduction to Mechanics. McGraw-Hill. pp. 133–134. ISBN0-07-035048-5. Recall that F = dP/dt was established for a system composed of a certain set of particles[. ... I]t is essential to deal with the same set of particles throughout the time interval[. ...] Consequently, the mass of the system can not change during the time of interest.